Logarithmic equations (II)

Statement

Solve the following equations after stating the existence conditions:

• (a) $\log_2(x)=3$
• (b) $1-2\log_3(x)=3$
• (c) $2\log_4(x)-3\log_2(x)=4$

Solution

(a) EC $x>0$. $x=2^3=8$.

(b) EC $x>0$. $-2\log_3 x=2 \Rightarrow \log_3 x=-1 \Rightarrow x=3^{-1}=\tfrac13$.

(c) EC $x>0$. Since $\log_4 x=\tfrac12\log_2 x$: $\log_2 x-3\log_2 x=-2\log_2 x=4 \Rightarrow \log_2 x=-2 \Rightarrow x=2^{-2}=\tfrac14$.

$\boxed{x_a=8;\quad x_b=\tfrac13;\quad x_c=\tfrac14}$